Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $23.6$ years; the standard deviation is $5.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living less than $35$ years.
Explanation: $23.6$ $17.9$ $29.3$ $12.2$ $35$ $6.5$ $40.7$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $23.6$ years. We know the standard deviation is $5.7$ years, so one standard deviation below the mean is $17.9$ years and one standard deviation above the mean is $29.3$ years. Two standard deviations below the mean is $12.2$ years and two standard deviations above the mean is $35$ years. Three standard deviations below the mean is $6.5$ years and three standard deviations above the mean is $40.7$ years. We are interested in the probability of a gorilla living less than $35$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the gorillas will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $12.2$ years and the other half $({2.5\%})$ will live longer than $35$ years. The probability of a particular gorilla living less than $35$ years is ${95\%} + {2.5\%}$, or $97.5\%$.